Chapter 31.3 - Practical Applications of the Cartesian Plane

In the previous section we saw some solved examples which demonstrated the method of plotting points. In this section we will see some practical applications of the Cartesian plane.

Consider the rectangle ABCD in fig.31.18 below.  

Fig.31.18

The following data is given:
• It is drawn on a Cartesian plane.
    ♦ The scale is: 1 unit = 1 cm for both the axes
• Length and width of the rectangle ABCD are 4 cm and 2 cm respectively. 
• The coordinates of it's lower left corner A is (3,2)
• The length AB is parallel to the x-axis 
■ We are asked to find the coordinates of the other three corners B, C and D
Solution:
1. To find the coordinates of B:
(i) Given that AB is parallel to the x-axis. So every point on AB will be at the same distance from the x-axis. 
• So every point on AB will have the same y coordinate as A. 
• The corner B is a point on AB. So B will have the same y coordinate as A. 
• Thus we get: y coordinate of B is 2
(ii) Length of AB is given as 4 cm. So B is at a horizontal distance of 4 cm from A
• A is at a horizontal distance of 3 units from the y-axis. 
• So B will be at a horizontal distance of (3+4) = 7 units from the y-axis. 
• Thus we get: x coordinate of B is 7
(iii) From (i) and (ii) we get: The coordinates of B are: (7,2)
2. To find the coordinates of C:
(i) BC is parallel to the y-axis. So every point on BC will be at the same distance from the y-axis. 
• So every point on BC will have the same x coordinate as B. 
• The corner C is a point on BC. 
• So C will have the same x coordinate as B. 
• Thus we get: x coordinate of C is 7
(ii) Length of BC is given as 2 cm. So C is at a vertical distance of 2 cm from B
• B is at a vertical distance of 2 units from the x-axis. 
• So C will be at a vertical distance of (2+2) = 4 units from the x-axis. 
• Thus we get: y coordinate of C is 4
(iii) From (i) and (ii) we get: The coordinates of C are: (7,4)
3. To find the coordinates of D:
(i) AD is parallel to the y-axis. So every point on AD will be at the same distance from the y-axis. 
• So every point on AD will have the same x coordinate as A. 
• The corner D is a point on AD. 
• So D will have the same x coordinate as A. 
• Thus we get: x coordinate of D is 3
(ii) Length of AD is given as 2 cm. So D is at a vertical distance of 2 cm from A
• A is at a vertical distance of 2 units from the x-axis. 
• So D will be at a vertical distance of (2+2) = 4 units from the x-axis. 
• Thus we get: y coordinate of D is 4
(iii) From (i) and (ii) we get: The coordinates of D are: (3,4)
The newly obtained coordinates are shown in the fig.31.19 below:

Fig.31.19

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We can see two interesting facts from the above problem:

Fact 1:
• Consider the line AB
    ♦ It is parallel to the x-axis
    ♦ If extended towards the left, it will pass through the point 2 on the y-axis      
• All points on AB will have the same y coordinate '2'
• Consider the line DC
    ♦ It is parallel to the x-axis
    ♦ If extended towards the left, it will pass through the point 4 on the y-axis      
• All points on DC will have the same y coordinate '4'
■ From this we can write:
• If a line is parallel to the x-axis, and if (when extended) it passes through a point 'm' on the y-axis, all points on that line will have the y-coordinate as 'm'
• In fact, that line can be represented by the equation: y = m
Fact 2:
• Consider the line AD
    ♦ It is parallel to the y-axis
    ♦ If extended towards the bottom, it will pass through the point 3 on the x-axis      
• All points on AD will have the same x coordinate '3'
• Consider the line BC
    ♦ It is parallel to the y-axis
    ♦ If extended towards the bottom, it will pass through the point 7 on the x-axis      
• All points on BC will have the same x coordinate '7'
■ From this we can write:
• If a line is parallel to the y-axis, and if (when extended) it passes through a point 'n' on the x-axis, all points on that line will have the x-coordinate as 'n'
• In fact, that line can be represented by the equation: x = n
Some examples are shown in the fig.31.20 below:

Fig.31.20

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Position of the axes

Consider the blue and green rectangles in fig.31.21 below:

Fig.31.21

1. The coordinates of all the corners are given. After some simple calculations, we will be able to find two facts:
• The length of both the rectangles are 4 units
• The width of both the rectangles are 2 units
2. Now consider the fig.31.22 below:

Fig.31.22

• The position of 'O' has been changed. The origin is now at the lower left corner of the green rectangle. This is also the upper left corner of the blue rectangle
3. After some simple calculations, we will be able to find two facts:
• The length of both the rectangles are 4 units
• The width of both the rectangles are 2 units
The reader may do the calculations in his/her own notebooks and verify the above facts
4. So we find that, changing the position of the coordinate axes do not have any effect on the properties of the 'objects'. The only change is to the 'coordinates of points'.
5. That means, we can place the axes at any convenient point that we like. But the following conditions should be satisfied:
• The axes must be perpendicular to each other
• Once the position of axes is fixed, it must not be changed

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Now we will see some solved examples:

Solved example 31.8
Find the coordinates of the other three vertices of the rectangle OABC in the fig.31.23(a) below

Fig.31.23

Solution:
1. Coordinates of 'O' are obviously (0,0)
2. Coordinates of A:
(i) Given that OABC is a rectangle. 
• So AB is parallel to OC. 
• But OC lies on the y-axis. So AB is parallel to the y-axis
• So any point on AB will have the same x coordinate as B
• Thus the x coordinate of A is 4
(ii) A lies on the x-axis
• Any point on the x-axis will have the y coordinate as '0'
(iii) From (i) and (ii) we get: coordinates of A are (4,0)
3. Coordinates of C:
(i) Given that OABC is a rectangle. 
• So BC is parallel to OA. 
• But OA lies on the x-axis. So BC is parallel to the x-axis
• So any point on BC will have the same y coordinate as B
• Thus the y coordinate of C is 3
(ii) C lies on the y-axis
• Any point on the y-axis will have the x coordinate as '0'
(iii) From (i) and (ii) we get: coordinates of C are (0,3)
• The newly calculated coordinates are shown in fig.31.23(b) above

Solved example 31.9
Find the coordinates of the other three vertices A, B and D of the rectangle in the fig.31.24(a) below. Given that, the origin O is at the centre of the rectangle and the sides of the rectangle are parallel to the axes.

Fig.31.24

Solution:
1. Coordinates of B:
(i) Given that BC is parallel to y-axis
• So any point on BC will have the same x coordinate as B
• Thus the x coordinate of B is 3
(ii) The x-axis will pass through the midpoint of BC. So we get:
• Distance of C from x-axis = Distance of B from x-axis
• But distance of C from x-axis = it's y coordinate = 2
• So distance of B from the x axis = 2
• But B is in the fourth quadrant. In the fourth quadrant, the y-coordinate should be given -ve sign
• So the y coordinate of B is -2
(iii) From (i) and (ii) we get: coordinates of B are (3,-2)
2. Coordinates of A:
(i) AB is parallel to x-axis
• So any point on AB will have the same y coordinate as B
• Thus the y coordinate of A is -2
(ii) The y-axis will pass through the midpoint of AB. So we get:
• Distance of B from y-axis = Distance of A from y-axis
• But distance of B from y-axis = it's x coordinate = 3
• So distance of A from the x axis = 3
• But A is in the third quadrant. In the third quadrant, the x-coordinate should be given -ve sign
• So the x coordinate of A is -3
(iii) From (i) and (ii) we get: coordinates of A are (-3,-2)
3. Coordinates of D:
(i) CD is parallel to x-axis
• So any point on CD will have the same y coordinate as C
• Thus the y coordinate of D is 2
(ii) The y-axis will pass through the midpoint of CD. So we get:
• Distance of C from y-axis = Distance of D from y-axis
• But distance of C from y-axis = it's x coordinate = 3
• So distance of D from the x axis = 3
• But D is in the second quadrant. In the second quadrant, the x-coordinate should be given -ve sign
• So the x coordinate of D is -3
(iii) From (i) and (ii) we get: coordinates of D are (-3,2)

Solved example 31.10
The isosceles triangle ABC shown in fig.31.25(a) has the base AB of 3 units and height 4 units. 

Fig.31.25

What are the coordinates of the vertices A, B and C?
Solution:
1. The x-axis is passing through the base
• The y-axis, which is always perpendicular to the x-axis, is passing through the apex
• So the altitude of the given triangle lies on the y-axis
• Thus we get: Distance of C from the x-axis = altitude of the triangle = 4 units
• So y coordinate of C = 4 (This is +ve 4 because it is on OY, the positive side of y-axis)
2. The x coordinate of any point on the y axis is 0
3. From (1) and (2) we get: Coordinates of C are: (0,4)
4. Since y axis is the altitude, it bisects the base AB perpendicularly. So OA = OB = ³⁄₂ = 1.5 units
• Now we can write the coordinates of A and B:
5. Coordinates of A:
(i) A is at a distance of 1.5 units from y-axis. So the x coordinate of A is -1.5 (-ve sign is given because A lies on OX' which is the negative side of the x-axis)
(ii) The y coordinate of any point on the x-axis is 0
(iii) From (i) and (ii) we get: Coordinates of A are (-1.5,0)
6. Coordinates of B:
(i) B is at a distance of 1.5 units from y-axis. So the x coordinate of B is +1.5 (+ve sign is given because B lies on OX which is the positive side of the x-axis)
(ii) The y coordinate of any point on the x-axis is 0
(iii) From (i) and (ii) we get: Coordinates of B are (1.5,0)
7. The results are shown in fig,31.25(b)

Solved example 31.11
An equilateral triangle OAB having side 4 units is placed in the Cartesian plane as shown in the fig.31.26(a) below:

Fig.31.26

Find the coordinates of the vertices A and B
Solution:
1. Coordinates of A:
(i) A lies on the x-axis and is at a distance of 4 units from O
• So it's x coordinate is +4 (∵ it lies on the positive side OX)
(ii) All points on the x-axis will have y coordinate as 0. 
(iii) From (i) and (ii) we get: coordinates of A are (4,0)
2. Coordinates of B:
(i) Consider fig.31.26(b). Two white dashed lines, one horizontal and the other vertical, pass through B
(ii) We want the following two information:
• The distance of the horizontal dashed line from the x-axis
• The distance of the vertical dashed line from the y-axis
(iii) First consider the vertical dashed line. 
• Since the triangle is equilateral, the vertical dashed line will pass through the midpoint of the base OA
• That means, the vertical dashed line is at a distance of 2 units from the y-axis
• x coordinate of any point on this vertical dashed line will be 2
• So x coordinate of B is 2
(iv) Now consider the horizontal dashed line. 
• The distance of this line from the x-axis is same as the height of the given equilateral triangle
• We have: Height of any equilateral triangle of side s = ^(√3)s⁄₂ (Details here)
• So the height of our equilateral triangle OAB = ^{(√3)× 4}⁄₂ = 2√3
• So distance of the horizontal dashed line from the x-axis is 2√3
• y coordinate of any point on this horizontal dashed line will be 2√3
• So y coordinate of B is 2√3
(v) From (iii) and (iv) we get: Coordinates of B are: (2,2√3)

Solved example 31.12
A parallelogram having longer side 6 units and shorter side 4 units is placed in the Cartesian plane as shown in the fig.31.27(a) below:

Fig.31.27

The included angle is 60^o. Find the coordinates of the vertices A, B and C
Solution:
1. Coordinates of A:
(i) A lies on the x-axis and is at a distance of 6 units from O
• So it's x coordinate is +6 (∵ it lies on the positive side OX)
(ii) All points on the x-axis will have y coordinate as 0. 
(iii) From (i) and (ii) we get: coordinates of A are (6,0)

2. Coordinates of B:
(i) Consider fig.31.27(b). Two white dashed lines, one horizontal and the other vertical, pass through B
(ii) We want the following two information:
• The distance of the horizontal dashed line from the x-axis
• The distance of the vertical dashed line from the y-axis

(iii) First consider the vertical dashed line.
It meets the x-axis at B'. So we have a right triangle AB'B
• It is a 30^o, 60^o triangle. We have seen the details of such triangles here.
• The hypotenuse will be 2 times the smallest side
• So AB = 2×AB' ⟹AB' = ^AB⁄₂ = ⁴⁄₂ = 2 units
• Then distance OB' = (OA + AB') = (6+2) = 8 units
• That means, the vertical dashed line is at a distance of 8 units from the y-axis
• x coordinate of any point on this vertical dashed line will be 8
• So x coordinate of B is 8
(iv) Now consider the horizontal dashed line
• In the right triangle AB'B, the altitude will be √3 times the smallest side
• So BB' = √3×AB' = 2√3
• So distance of the horizontal dashed line from the x-axis is 2√3
• y coordinate of any point on this horizontal dashed line will be 2√3
• So y coordinate of B is 2√3
(v) From (iii) and (iv) we get: Coordinates of B are: (8,2√3)
3. Coordinates of C:
(i) C is on the same horizontal line as B. 
• So the y coordinate of C will be same as that of B, which is 2√3
(ii) The horizontal dashed line meets the y-axis at C'
• By properties of a parallelogram, C'C will be equal to AB' which is 2 units
• So the x coordinate of C is 2
(iii) From (i) and (i) we get: Coordinates of C are: (2,2√3)

Solved example 31.13
In the fig.31.28(a) below, a large trapezium is made up of four equal trapeziums. 

Fig.31.28

Find the coordinates of the vertices of all the trapeziums
Solution:
1. Consider the red and green trapeziums at the bottom
• Two equal bases give a distance of 8 units. So each of them is (⁸⁄₂) = 4 units
2. Both the bases lie on the x-axis. So we get:
• Bottom right vertex of red is (4,0)
• Bottom right vertex of green is (8,0)
3. The height of the total large trapezium is the base of the magenta trapezium
• But base of magenta = base of red = 4 units
• So top right of magenta will have a y coordinate of 4
• It's x coordinate will be: (base of red + base of green) = (4+4) = 8
• So the coordinates at the top right of magenta is (8,4)
4. y coordinate at the top left of magenta will be same as that at it's top right which is 4
• Height of red + height of blue = total height = 4
• So height of red = height of blue = height of each = ⁴⁄₂ = 2 units
• x coordinate of top left of magenta = (x coordinate of top right of magenta - 2) = (8-2) = 6
• So the top left of magenta is (6,4)
■ In this way all other vertices can be calculated. They are shown in fig.b
Note:
• Top of red + top of green = base of blue = 4
• So top of red = top of green = top of each = ⁴⁄₂ = 2 units

Solved example 31.14
All the rectangles in fig.31.29 below have sides parallel to the axes. 

Fig.31.29

Find the coordinates of the remaining vertices of each.
Solution:
Fig.a:
1. To find the coordinates of B:
(i) Given that AB is parallel to the x-axis. So every point on AB will be at the same distance from the x-axis. 
• So every point on AB will have the same y coordinate as A. 
• The corner B is a point on AB. So B will have the same y coordinate as A. 
• Thus we get: y coordinate of B is 3
(ii) Given that BC is parallel to the y-axis. So every point on BC will be at the same distance from the y-axis. 
• So every point on BC will have the same x coordinate as A. 
• The corner B is a point on BC. So B will have the same x coordinate as C. 
• Thus we get: x coordinate of B is 2
(iii) From (i) and (ii) we get: The coordinates of B are: (2,3)
2. To find the coordinates of D:
(i) Given that CD is parallel to the x-axis. So every point on CD will be at the same distance from the x-axis. 
• So every point on CD will have the same y coordinate as C. 
• The corner D is a point on CD. So D will have the same y coordinate as C. 
• Thus we get: y coordinate of D is 4
(ii) Given that AD is parallel to the y-axis. So every point on AD will be at the same distance from the y-axis. 
• So every point on AD will have the same x coordinate as A. 
• The corner D is a point on AD. So D will have the same x coordinate as A. 
• Thus we get: x coordinate of D is -2
(iii) From (i) and (ii) we get: The coordinates of B are: (-2,4)

Fig.b:
1. To find the coordinates of A:
(i) Given that AB is parallel to the x-axis. So every point on AB will be at the same distance from the x-axis. 
• So every point on AB will have the same y coordinate as B. 
• The corner A is a point on AB. So A will have the same y coordinate as B. 
• Thus we get: y coordinate of A is -4
(ii) Given that AD is parallel to the y-axis. So every point on AD will be at the same distance from the y-axis. 
• So every point on AD will have the same x coordinate as D. 
• The corner A is a point on AD. So A will have the same x coordinate as D. 
• Thus we get: x coordinate of A is -1
(iii) From (i) and (ii) we get: The coordinates of B are: (-1,-4)
2. To find the coordinates of C:
(i) Given that CD is parallel to the x-axis. So every point on CD will be at the same distance from the x-axis. 
• So every point on CD will have the same y coordinate as D. 
• The corner C is a point on CD. So C will have the same y coordinate as D. 
• Thus we get: y coordinate of C is -2
(ii) Given that BC is parallel to the y-axis. So every point on BC will be at the same distance from the y-axis. 
• So every point on BC will have the same x coordinate as B. 
• The corner C is a point on BC. So C will have the same x coordinate as B. 
• Thus we get: x coordinate of C is 2
(iii) From (i) and (ii) we get: The coordinates of C are: (2,-2)

Fig.c:
1. To find the coordinates of B:
(i) Given that AB is parallel to the x-axis. So every point on AB will be at the same distance from the x-axis. 
• So every point on AB will have the same y coordinate as A. 
• The corner B is a point on AB. So B will have the same y coordinate as A. 
• Thus we get: y coordinate of B is 3
(ii) Given that BC is parallel to the y-axis. So every point on BC will be at the same distance from the y-axis. 
• So every point on BC will have the same x coordinate as A. 
• The corner B is a point on BC. So B will have the same x coordinate as C. 
• Thus we get: x coordinate of B is 2
(iii) From (i) and (ii) we get: The coordinates of B are: (2,3)
2. To find the coordinates of D:
(i) Given that CD is parallel to the x-axis. So every point on CD will be at the same distance from the x-axis. 
• So every point on CD will have the same y coordinate as C. 
• The corner D is a point on CD. So D will have the same y coordinate as C. 
• Thus we get: y coordinate of D is 6
(ii) Given that AD is parallel to the y-axis. So every point on AD will be at the same distance from the y-axis. 
• So every point on AD will have the same x coordinate as A. 
• The corner D is a point on AD. So D will have the same x coordinate as A. 
• Thus we get: x coordinate of D is -1
(iii) From (i) and (ii) we get: The coordinates of B are: (-1,6)
The completed fig.31.30 is shown below:

Fig.31.30

Solved example 31.15
Given below are four pairs of coordinates. Each pair contains coordinates of any two corners of rectangles whose sides are parallel to the axes. In each pair, mark each coordinate as top-left, top-right, bottom-left or bottom-right. Mark them with out drawing axes. After marking, find the other coordinates also.
(i) (3,5), (7,8)
(ii) (6,2), (5,4)
(iii) (-3,5), (-7,1)
(iv) (-1,-2), (-5,-4)
Solution:
Case (i):
1. Consider the x coordinates: 3 and 7
(i) 7 is greater than 3. So (7,8) is on the right side of (3,5)
(ii) Thus the possibilities for (3,5) are: left top or left bottom
(iii) Also possibilities for (7,8) are: right top or right bottom
2. Consider the y coordinates: 5 and 8
(i) 8 is greater than 5. So (3,5) is at bottom and (7,8) is at top  
(ii) Thus the possibilities for (3,5) are: left bottom or right bottom
(iii) Also possibilities for (7,8) are: left top or right top
3. Combining 1(ii) and 2(ii) we get: (3,5) is left bottom [Taking the common possibility. See the green highlighted lines]
4. Combining 1(iii) and 2(iii) we get: (7,8) is right top [Taking the common possibility. See the red highlighted lines]
This is shown in the fig.31.31 below:

Fig.31.31

Case (ii):
1. Consider the x coordinates: 6 and 5
(i) 6 is greater than 5. So (6,2) is on the right side of (5,4)
(ii) Thus the possibilities for (5,4) are: left top or left bottom
(iii) Also possibilities for (6,2) are: right top or right bottom
2. Consider the y coordinates: 2 and 4
(i) 5 is greater than 2. So (6,2) is at bottom and (5,4) is at top  
(ii) Thus the possibilities for (6,2) are: left bottom or right bottom
(iii) Also possibilities for (5,4) are: left top or right top
3. Combining 1(ii) and 2(iii) we get: (5,4) is left top [Taking the common possibility]
4. Combining 1(iii) and 2(ii) we get: (6,2) is right bottom [Taking the common possibility]
This is shown in the fig. above

Case (iii):
1. Consider the x coordinates: -3 and -7
(i) -3 is greater than -7. So (-3,5) is on the right side of (-7,1)
(ii) Thus the possibilities for (-7,1) are: left top or left bottom
(iii) Also possibilities for (-3,-5) are: right top or right bottom
2. Consider the y coordinates: 5 and 1
(i) 5 is greater than 1. So (-7,1) is at bottom and (-3,5) is at top  
(ii) Thus the possibilities for (-7,1) are: left bottom or right bottom
(iii) Also possibilities for (-3,5) are: left top or right top
3. Combining 1(ii) and 2(ii) we get: (-7,1) is left bottom [Taking the common possibility]
4. Combining 1(iii) and 2(ii) we get: (-3,5) is right top [Taking the common possibility]
This is shown in the fig. above

Case (iv):
1. Consider the x coordinates: -1 and -5
(i) -1 is greater than -5. So (-1,-2) is on the right side of (-5,-4)
(ii) Thus the possibilities for (-5,4) are: left top or left bottom
(iii) Also possibilities for (-1,-2) are: right top or right bottom
2. Consider the y coordinates: -2 and -4
(i) -2 is greater than -4. So (-5,-4) is at bottom and (-1,-2) is at top  
(ii) Thus the possibilities for (-5,-4) are: left bottom or right bottom
(iii) Also possibilities for (-1,-2) are: left top or right top
3. Combining 1(ii) and 2(ii) we get: (-5,-4) is left bottom [Taking the common possibility]
4. Combining 1(iii) and 2(ii) we get: (-1,-2) is right top [Taking the common possibility]
This is shown in the fig. above
Part 2:
The remaining coordinates of each rectangle can be calculated by using the procedure that we saw in the previous example. The completed figs are shown below:

Fig.31.32 

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In the next section we will see how to calculate the distance between any two points when their coordinates are given.

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Chapter 31.2 - Plotting points when coordinates are given

In the previous section we saw how to find the coordinates of points in a Cartesian plane. In this section we will see the reverse process. That is., we will be given coordinates of a point. We must locate them on the Cartesian plane. The process of marking points on the Cartesian plane, when their coordinates are given, is called Plotting the points.

Let us see an example:
Given: Coordinates of a point P are (5,3). Plot the point on the Cartesian plane
Solution:
1. The most efficient way for plotting points is to use a graph paper.
2. So, on a graph paper, draw the two coordinate axes: X'X and Y'Y
3. Mark their point of intersection as the origin O
4. Mark the points on the axes choosing a suitable scale. For this problem, '1 cm = 1 unit' will be appropriate. (See details for scale here)
5. Now consider the coordinates. We have '5' as the x-coordinate.
• So the required point will be at a distance of 5 units from the y axis. 
6. Draw a line parallel to the y axis. This line should pass through the '5' on the x-axis.
• But on the x axis, there is '5' and '-5'. We must take '5'.
• This is because, 'positive 5' indicates that it is on the positive side of the x-axis. That is., OX 
• This line is shown in red colour in fig.31.12 below:
7. It is clear that, the required point lies somewhere on the red line. But where?
• To find that, we consider the y-coordinate. It is '3'
• So the required point is at a distance of 3 units from the x-axis. Also, since it is 'positive 3', the point lies on the positive side of the y axis. That is., OY
8. So draw a line parallel to the x-axis through '3' on OY. This is the green horizontal line in fig.31.12
• It is clear that the required point lies somewhere on the green line. But where?       
9. Consider the point of intersection of the red and green lines. That point satisfies both the conditions:
• It is at a distance of 3 5 units from the y-axis
• It is at a distance of 3 units from the x-axis
10. So the point of intersection is our required point P
■ Note that, if we use a graph paper, we will not need to draw the red and green lines because, they will be already present. All we would need to do then, is 'count the number of units' along the axes. This is shown in the fig.31.13 below:

Fig.31.13

Solved example 31.4
Locate the points (5,0), (0,5), (2,5), (5,2), (-3,5), (-3,-5), (5,-3) and (6,1) in the Cartesian plane
Solution:
• The most efficient way for plotting points is to use a graph paper.
• So, on a graph paper, draw the two coordinate axes: X'X and Y'Y
• Mark their point of intersection as the origin O
• Mark the points on the axes choosing a suitable scale. For this problem, '1 cm = 1 unit' will be appropriate.
A. The given coordinates are (5,0)
1. Draw a vertical line through '5' in OX (OX is used since '5' is positive)
2. Draw a horizontal line through '0'. 
• '0' in which one? OX, OX', OY or OY' ?
Ans: There is only one zero. That is at the origin 'O'
• So draw a horizontal through the origin 'O'
• But the horizontal through O is the x-axis
3. So the required lines are:
• The vertical line through '5' in OX
• The x-axis
4. The point of intersection of these two lines will have the coordinates: (5,0)
See fig.31.14 below

Fig.31.14

B. The given coordinates are (0,5)
1. Draw a vertical line through'0'. 
• '0' in which one? OX, OX', OY or OY' ?
Ans: There is only one zero. That is at the origin 'O'
• So draw a vertical line through the origin 'O'
• But the vertical line through O is the y-axis
2. Draw a horizontal line through '5' in OY (OY is used since '5' is positive)
3. So the required lines are:
• The y-axis
• The horizontal line through '5' in OY
4. The point of intersection of these two lines will have the coordinates: (0,5)
See fig.31.14 above
C. The given coordinates are (2,5)
1. Draw a vertical line through '2' in OX (OX is used since '2' is positive)
2. Draw a horizontal line through '5' in OY (OY is used since '5' is positive)
3. The point of intersection of these two lines will have the coordinates: (2,5)
See fig.31.14 above
D. The given coordinates are (5,2)
1. Draw a vertical line through '5' in OX (OX is used since '5' is positive)
2. Draw a horizontal line through '2' in OY (OY is used since '2' is positive)
3. The point of intersection of these two lines will have the coordinates: (5,2)
See fig.31.14 above
E. The given coordinates are (-3,5)
1. Draw a vertical line through '-3' in OX' (OX' is used since '-3' is negative)
2. Draw a horizontal line through '5' in OY (OY is used since '5' is positive)
3. The point of intersection of these two lines will have the coordinates: (-3,5)
See fig.31.14 above
F. The given coordinates are (-3,-5)
1. Draw a vertical line through '-3' in OX' (OX' is used since '-3' is negative)
2. Draw a horizontal line through '-5' in OY' (OY' is used since '-5' is positive)
3. The point of intersection of these two lines will have the coordinates: (-3,-5)
See fig.31.14 above
G. The given coordinates are (5,-3)
1. Draw a vertical line through '5' in OX (OX is used since '5' is positive)
2. Draw a horizontal line through '-3' in OY' (OY' is used since '-3' is positive)
3. The point of intersection of these two lines will have the coordinates: (5,-3)
See fig.31.14 above
H. The given coordinates are (6,1)
1. Draw a vertical line through '6' in OX (OX is used since '6' is positive)
2. Draw a horizontal line through '1' in OY (OY is used since '1' is positive)
3. The point of intersection of these two lines will have the coordinates: (6,1)

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• In the above example, we can note the following points:
    ♦ (5,0) and (0,5) are not at the same position
    ♦ (-3,5) and (5,-3) are not at the same position
• Several such examples can be shown in a Cartesian plane
■ So in general, we can write:
IF x ≠ y THEN, (x,y) and (y,x) are not at the same position
• That means, we cannot interchange x and y. 
• The order of x and y is important in (x,y)
• So (x,y) is called an ordered pair
■ IF x ≠ y, THEN, [ordered pair (x,y)] ≠ [ordered pair (y,x)]
■ IF x = y, THEN, [ordered pair (x,y) = ordered pair (y,x)]

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Solved example 31.5
Plot the following ordered pairs (x,y) of numbers as points in the Cartesian plane.

x	-3	0	-1	4	2
y	7	-3.5	-3	4	-3

Use scale 1 cm = 1 unit on the axes
Solution:
The points are plotted in the fig.31.15 below:

Fig.31.15

The reader may make his/her own plot and compare with the above fig.
Solved example 31.6
In which quadrant or on which axis do each of the points (-2,4), (3,-1), (-1,0), (1,2) and (-3,-5) lie? Verify your answers by locating them on the Cartesian plane
Solution:
We can use the following rules:
• Points in the first quadrant will be in the form: (+,+) 
• Points in the second quadrant will be in the form: (-,+) 
• Points in the third quadrant will be in the form: (-,-) 
• Points in the fourth quadrant will be in the form: (+,-) 
• Points in the x-axis will be in the form: (±x,0) 
• Points in the y-axis will be in the form: (0, ±y)
1. (-2,4) will lie in the second quadrant 
2. (3,-1) will lie in the fourth quadrant 
3. (-1,0) will lie in the x-axis
4. (1,2) will lie in the first quadrant 
5. (-3,-5) will lie in the third quadrant
The plot is shown in the fig.31.16 below:

Fig.31.16

Solved example 31.7
Plot the following ordered pairs (x,y) of numbers as points in the Cartesian plane.

x	-2	-1	0	1	3
y	8	7	-1.25	3	-1

Use scale 1 cm = 1 unit on the axes
Solution:
The points are plotted in the fig.31.17 below:

Fig.31.17

The reader may make his/her own plot and compare with the above fig.

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In the next section we will see some very simple practical applications of the Cartesian plane.

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Chapter 31 - Coordinate Geometry

In the previous section we completed a discussion on Trigonometry. In this section we will see Coordinate Geometry. We saw some basic details about graphs in chapter 2. Here we will have a more detailed study.

We have studied about the number lines also in an earlier chapter 22.  Following are the basic features of a number line:
• The distances are marked from a fixed point
    ♦ This fixed point is called the origin
• The distances are marked in equal units
    ♦ If 1 unit from the origin represents the number 1, then 3 units from origin will represent the number 3
    ♦ In the same way, the number 'r' will be 'r units' away from the origin 
• The distances are marked positively in one direction and negatively in the opposite direction
    ♦ The number 'r' will be 'r units' away from the origin in the positive direction
    ♦ The number '-r' will be 'r units' away from the origin in the negative direction
• An example is shown in the fig.31.1 below:

Fig.31.1

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• Consider two such number lines. Place them perpendicular to each other. 

    ♦ But there are infinite possibilities to place 'two lines perpendicular to each other'. The xxx fig.31.2 below shows some of them:

Fig.31.2

• All the three arrangements in the above fig.31.2 satisfies one condition:

    ♦ The lines should be perpendicular to each other. 
• In this chapter, we will be using the arrangement in fig.c. That is:
    ♦ One line will be horizontal and the other line will be vertical

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Let us see how the numbering system can be applied to the two perpendicular lines:

1. Consider any simple horizontal number line as shown in the fig.31.3(a) below. Let us name it X'X
• The positive numbers are marked towards the right from the origin. 
• The negative numbers are marked towards the left from the origin. This is shown in fig.31.3(b)

Fig.31.3

2. Consider another number line. This new number line must be vertical. Let us name it Y'Y. See fig.31.3(b) 
• The positive numbers must be marked upwards from the origin. 
• The negative numbers must be marked downwards from the origin.
3. Now combine the two number lines. 
• The combination should be done in such a way that, the two lines cross each other at their origins.
4. The horizontal line X'X is called the x-axis
• The vertical line Y'Y is called the y-axis 
• The point where X'X and Y'Y cross is called the origin
    ♦ It is denoted by the letter 'O'
5. The positive numbers belonging to the x-axis lies on OX
    ♦ So OX is called the 'positive direction of the x-axis' 
• The negative numbers  belonging to the x-axis lies on OX'
    ♦ So OX' is called the 'negative direction of the x-axis'
• The positive numbers  belonging to the y-axis lies on OY
    ♦ So OY is called the 'positive direction of the y-axis' 
• The negative numbers  belonging to the y-axis lies on OY'
    ♦ So OY' is called the 'negative direction of the y-axis'

6. Consider the fig.31.4 below. Both the x-axis and y-axis are shown. 

Fig.31.4

• The two axes (plural of 'axis' is 'axes') divide the plane of the paper into four parts. 

• These four parts are called quadrants. 

    ♦ In other words, each one of the four parts can be called: A quadrant

(Recall that a polygon with four sides is called a quadrilateral)
7. Each quadrant is given a particular name. The procedure for naming is simple:
• Just number them as I, II, III and IV
• The numbering should be done in the anti clockwise direction starting from OX. So we get the following:
    ♦ Top right: Quadrant I, 
    ♦ Top left: Quadrant II
    ♦ Bottom left: Quadrant III 
    ♦ Bottom right: Quadrant IV
8. Thus the 'plane of the paper' consists of two axes and four quadrants
• This plane is known by three different names. We can use any one of them:
    ♦ The cartesian plane
    ♦ The coordinate plane
    ♦ The xy-plane
9. The axes are called coordinate axes

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Now we will see how the coordinate system can be used to specify the exact location any points on a plane. 

• Consider fig.31.5 below:

Fig.31.5

• Two points P and Q are marked on the cartesian plane. 

• We want to specify the exact locations of P and Q. For that, we use the following steps:

1. Drop the following perpendicular lines:
• perpendicular PM from P on to the x-axis (See fig.31.6 below) 
• perpendicular PN from P on to the y-axis
• perpendicular QR from Q on to the x-axis
• perpendicular QS from Q on to the y-axis

Fig.31.6

2. Dropping the perpendicular lines gives us the following information:
• The perpendicular distance of P from the y-axis is equal to PN. 
• We want the value of this PN. But PN is same as OM (∵ M is the foot of the perpendicular from P)
• We can easily see that OM is 4 units. So PN is 4 units. Thus we get:
• The point P is at a perpendicular distance of 4 units from the  y-axis
    ♦ This 4 units is measured along the positive direction of the x-axis
3. Working in a similar way we get these also:
• The point P is at a perpendicular distance of 3 units from the  x-axis
    ♦ This 3 units is measured along the positive direction of the y-axis
• The point Q is at a perpendicular distance of 6 units from the  y-axis
    ♦ This 6 units is measured along the negative direction of the x-axis
• The point Q is at a perpendicular distance of 2 units from the  x-axis
    ♦ This 2 units is measured along the positive direction of the y-axis
4. Now we have all the required information for specifying the location of the points.
• But there are three rules to be followed. 
• The rules will ensure that the 'final presentation of the results' made by all of us will be in the same form. 
• The 3 rules are:
(i) The x-coordinate of a point, is the perpendicular distance of that point from the y-axis
• This distance should be measured along the x-axis
    ♦ The x-coordinate thus obtained is positive if the measurement is done along OX  
    ♦ The x-coordinate thus obtained is negative if the measurement is done along OX'  
(ii) The y-coordinate of a point, is the perpendicular distance of that point from the x-axis
• This distance should be measured along the y-axis
    ♦ The y-coordinate thus obtained is positive if the measurement is done along OY  
    ♦ The y-coordinate thus obtained is negative if the measurement is done along OY'
(iii) Write the coordinates of a point inside brackets. They must be separated by a comma.
• The x-coordinate is written first and then the y-coordinate
    ♦ Another name for x-coordinate is abscissa
    ♦ Another name for y-coordinate is ordinate
5. Let us apply the rules to point P:
(i) Applying the first rule to find the x-coordinate:
• Perpendicular distance of P from the y-axis (measured along the x-axis) is '4 units'
    ♦ It is measured along OX. So it is positive
    ♦ Thus the x-coordinate of P is '4'
(ii) Applying the second rule to find the y-coordinate:
• Perpendicular distance of P from the x-axis (measured along the y-axis) is '3 units'
    ♦ It is measured along OY. So it is positive
    ♦ Thus the y-coordinate of P is '3'
(iii) Applying the third rule to write down the coordinates:
• x-coordinate first and then the y-coordinate. Inside brackets and separated by comma. We get:
■ The coordinates of P are: (4,3)
6. Let us apply the rules to point Q:
(i) Applying the first rule to find the x-coordinate:
• Perpendicular distance of Q from the y-axis (measured along the x-axis) is '6 units'
    ♦ It is measured along OX'. So it is negative
    ♦ Thus the x-coordinate of P is '-6'
(ii) Applying the second rule to find the y-coordinate:
• Perpendicular distance of Q from the x-axis (measured along the y-axis) is '2 units'
    ♦ It is measured along OY'. So it is negative
    ♦ Thus the y-coordinate of Q is '-2'
(iii) Applying the third rule to write down the coordinates:
• x-coordinate first and then the y-coordinate. Inside brackets and separated by comma. We get:
■ The coordinates of Q are: (-6,-2)
7. In the above example, we drew perpendicular lines PM, PN, QR and QS. If the points P and Q are marked on a graph paper, we will not need to draw those lines. This is shown in the fig.31.7 below:

Fig.31.7

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In the next section we will see some solved examples.

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